Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $q = \dfrac{t - 8}{5t^2 - 70t + 240} \div \dfrac{-t - 5}{2t^2 - 6t - 80} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{t - 8}{5t^2 - 70t + 240} \times \dfrac{2t^2 - 6t - 80}{-t - 5} $ First factor out any common factors. $q = \dfrac{t - 8}{5(t^2 - 14t + 48)} \times \dfrac{2(t^2 - 3t - 40)}{-(t + 5)} $ Then factor the quadratic expressions. $q = \dfrac {t - 8} {5(t - 8)(t - 6)} \times \dfrac {2(t - 8)(t + 5)} {-(t + 5)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {(t - 8) \times 2(t - 8)(t + 5) } { 5(t - 8)(t - 6) \times -(t + 5)} $ $q = \dfrac {2(t - 8)(t + 5)(t - 8)} {-5(t - 8)(t - 6)(t + 5)} $ Notice that $(t - 8)$ and $(t + 5)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {2\cancel{(t - 8)}(t + 5)(t - 8)} {-5\cancel{(t - 8)}(t - 6)(t + 5)} $ We are dividing by $t - 8$ , so $t - 8 \neq 0$ Therefore, $t \neq 8$ $q = \dfrac {2\cancel{(t - 8)}\cancel{(t + 5)}(t - 8)} {-5\cancel{(t - 8)}(t - 6)\cancel{(t + 5)}} $ We are dividing by $t + 5$ , so $t + 5 \neq 0$ Therefore, $t \neq -5$ $q = \dfrac {2(t - 8)} {-5(t - 6)} $ $ q = \dfrac{-2(t - 8)}{5(t - 6)}; t \neq 8; t \neq -5 $